Hallo allemaal,


Please have a look at this code:

What value should A contain? I expected $20 but it appears to be $30. Running the ML in VICE shows that A contains $30 but the P register $20. Not sure if the monitor is too blame, I created this ML:

Running the ML the second 'A' turns into a Zero, confirming A contains $30.

The big question: WHY ???

Many thanks for an explanation!

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The PHP always pushes a '1' BRK bit (bit 4, value $10). The only way to see if and ISR was entered by a BRK instruction is to pull the stacked record of P that was put there by the interrupt itself. This was covered in this topic.

Thank you!

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Just for the record: you have to pull the saved P into A using PLA (or load it into a register by other means) - you can't use PLP because P as a register only has 6 useful bits. (The value saved by an interrupt has 7 useful bits, but the transistors just aren't there in the P register itself.)

Also note that the W65C816S has a different behavior when operation in native mode. In this case, the value pushed to the stack by PHP and subsequently retrieved by PLA (PLX, or PLY) does not change. Hence:

returns $00 in .A.

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x86?  We ain't got no x86.  We don't NEED no stinking x86!

I think a clear way to understand this is to realize that the B flag is not a flag at all but only the logical OR of the I flag and the /IRQ pin. The bit pulled from the stack into the P register is just discarded.
I was suspecting this, and after digging in the 6502 schematic from http://impulzus.sch.bme.hu/6502/letolt.php3, I found it is true. I also found that this signal can clear the Instruction Register: In fact, fetching a BRK op-code instead of the byte present on the data bus when an interrupt is accepted. This can also explain why the BRK instruction is ignored if an interrupt is accepted at the same time (a well known bug of the NMOS 6502)


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